Prove the identity
\((\sec\theta - \tan\theta)^2 \equiv \dfrac{1 - \sin\theta}{1 + \sin\theta}\)
Solution
\(\sec\theta - \tan\theta = \dfrac{1}{\cos\theta} - \dfrac{\sin\theta}{\cos\theta}\)
\(= \dfrac{1 - \sin\theta}{\cos\theta}\)
\((\sec\theta - \tan\theta)^2 = \dfrac{(1 - \sin\theta)^2}{\cos^2\theta}\)
\(= \dfrac{(1 - \sin\theta)^2}{1 - \sin^2\theta}\) \(\;\) [since \(\cos^2\theta = 1 - \sin^2\theta\)]
\(= \dfrac{(1 - \sin\theta)^2}{(1 - \sin\theta)(1 + \sin\theta)}\)
\(= \dfrac{1 - \sin\theta}{1 + \sin\theta}\)
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